Absolute values and Subsets

[Sundelights]

I know that the absolute value of lets say -3 is 3 and the absolute value of 3 is 3, but i am having problems with equations like this:

4 + |r +2| =7

|2x -7| -1 > 0

For the first one would that be r= 1 and -1
For the second one would x =4 and -4

Also I am having problems with sets for instance;

Let N={x| x is a multiple of 2} and P={x| x is a multiple of 6}. Describe the intersection of N and P. I do not have any idea how to do that. I have an idea that it would be N and the union symbol and 6, 12, 18, and so forth, but I do not know how to write it. Could someone please help me. Thank you so much!!

 

[tkhunny]

I know that the absolute value of lets say -3 is 3 and the absolute value of 3 is 3, but i am having problems with equations like this:

4 + |r +2| =7

|2x -7| -1 > 0

For the first one would that be r= 1 and -1
For the second one would x =4 and -4

?? Why is it different. Absolute Value doesn't magically change its operation.

Get this idea stuck in your head.

|x| = x IF x >= 0 -- In other words, Absolute Value does nothing to nonnegative values.
|x| = -x IF x < 0 -- In other words, Absolute Value changes the sign of negative values.

See: |4| = 4 (nothing happened) and |-3| = 3 (sign changed)

Keep this in mind as your ponder |r+2|

|r+2| = r+2 if r+2 >= 0 or r >= -2
|r+2| = -(r+2) if r+2 < 0 or r < -2

Does this get us anywhere?

 

[pka]

I know that the absolute value of lets say -3 is 3 and the absolute value of 3 is 3, but i am having problems with equations like this:
4 + |r +2| =7
|2x -7| -1 > 0
For the first one would that be r= 1 and -1
For the second one would x =4 and -4

I think you need to go back to the basic definition of absolute value.
\(\displaystyle |x|\) is the distance the number \(\displaystyle x\) is from 0.
Thus \(\displaystyle |3|=|-3|=3\).
So if \(\displaystyle |t|=5\) then \(\displaystyle t=5\text{ or }t=-5\) because both of those are five units from zero.

If \(\displaystyle |2x+5|=1\) then \(\displaystyle 2x+5\) is a number that is one unit from zero.
Thus \(\displaystyle 2x+5=1\text{ or }2x+5=-1\).

Now let us look at \(\displaystyle |2x -7| -1 > 0\) or \(\displaystyle |2x -7| > 1\).
That says solve \(\displaystyle 2x-7>1\text{ or }2x-7<-1\).
Those have solutions \(\displaystyle x>4\text{ or }x<3\).